What does unique factorization exactly mean here? Given $a$ and $b$ are integers:
$$ 3\cdot 2 \cdot 5^2 \cdot 7^a = 2 \cdot 3^2 \cdot 7 \cdot 5^b $$
The answer is as follows:

Since both sides of this equation are integers and have unique factorizations, it follows that $a = 1$ and $b = 2$ is the only solution.

What does unique factorization exactly mean here? I looked up the word unique factorization on google but I don't really understand it. Can you explain it at a pre-calculus level? Also, what aspect of 'integers' makes the solution $a = 1$, and $b = 2$ ??
 A: Unique factorization means that there is only one way to factorize a positive integer (that means a positive whole number).
In basic terms, what this means is that if you write down a number as a product of primes, there is only one possible way to do that.
For example, if we write $52$ as a product of primes, we have
$52 = 2 \times 2 \times 13$.
There is no other way to write out $52$ as a product of primes. 
In your question, you have two expressions on either side of an equals sign, which means that they're the same number. The "answer" which you're talking about is telling you that given that you are writing this number down as a product of its prime factors, the powers of $7$ and powers of $5$ must be the same on both sides of the equation.
Does that make sense?
A: Every positive integer is either prime or composite or the number one.
If the number is composite then it can be factored as a product of smaller numbers.  Ex:  $120 = 12\cdot10$ or we could say $120 = 30\cdot4$ or we could say $120 = 6\cdot10\cdot2$.
If we factor a number into smaller composites and then factored those number we'd eventually factor it do to a factorization containing only primes.  Ex:  $120 = 12\cdot10 = (3\cdot4)\cdot(2\cdot5) = 3\cdot2\cdot2\cdot2\cdot5$  or we could say $120 = 30\cdot4 =(6\cdot5)\cdot(2\cdot2) = 2\cdot3\cdot5\cdot2\cdot2$ or $120=6\cdot10\cdot2 = 2\cdot3\cdot2\cdot5\cdot2$.
And if we have multiple occurrences of the same prime factor we can list them as a power.  Ex: $120 = 3\cdot2^3\cdot5$ or $120 = 2^3\cdot3\cdot5$ or $120 =3\cdot5\cdot2^3$.
Now perhaps the most anti-climatic conclusion you'll ever see.
No matter how you break a number down into primes you will always and up with the same primes and the same number of each of them.  Each number has only one unique way of being broken down into primes.
And that is what unique prime factorization means.
So if you have $3^2 \cdot 2 \cdot 5^2 \cdot 7^a = 2 \cdot 3^2 \cdot 7 \cdot 5^b$ then you have a number.  By the Left-Hand side we know it factors into: two $3$s, one $2$, two $5$s, and some $7$ but we don't know how many.  On the RHS we see that it factors down into: one $2$, two $3$s, one $7$, and some $5$s but we don't know how many.
But by the Unique prime factorizations we know the two lists of components must be exactly the same.  So how many $7$s are there?  Well, by the RHS we know there is one $7$. And how many $5$s are there?  Well, by the LHS we know there are two $5$s.
So that's that.  If $3^2 \cdot 2 \cdot 5^2 \cdot 7^a = 2 \cdot 3^2 \cdot 7 \cdot 5^b$ then $a =1$ and $b = 2$ and then number is $3^2 \cdot2\cdot5^2\cdot7 = 9\cdot2\cdot5^2\cdot7 = 3150$.

Note: The really hard way to do it:
$3^2 \cdot 2 \cdot 5^2 \cdot 7^a = 2 \cdot 3^2 \cdot 7 \cdot 5^b$  Divide both sides by $3^2\cdot2$
$5^2\cdot7^a = 7\cdot5^b$.  Divide both sides by $7$.
$5^2\cdot7^{a-1} = 5^b$.  Divide both sides by $5^2$.
$7^{a-1} = 5^{b-2}$.  
So.... the question is: what (integer) power can you raise $5$ to get a (integer) power of $7$ (or vice versa).  The answer is: $5$ and $7$ are prime; there isn't any such power. The only way this makes sense is if  $7^0  = 5^0 = 1$.  So $a-1 = b-2 =0$.
The point being: if you have ever told that  $3^a\cdot5^b = 3^c\cdot5^d$ and you are told $a$ and $b$ are  integers, then it should be immediately clear $a = c$ and $b = d$.
A: Of course, reordering factors is "cheating". The uniqueness referred to means that if you write all of the prime factors down, in order, then that is unique. This is often condensed by combining the repeats of a given prime factor into a single power of that prime. In that case, the uniqueness is of (1) the ordered list of distinct prime factors, together with (2) the corresponding list of positive exponents each such factor carries.
A: The numbers $2,3,5,7$ are all primes. 
Theorem: Any natural number greater than $1$ can be broken in a unique way into a product of primes. 
If there are repetitions in this product, this leads to powers ($2\times 2\times 2=2^3$, etc.) and the unicity from the theorem asserts then that these powers are unique, because they determine the number of repetitions for each prime.
For instance, it is impossible to have $2\times n=2\times 2\times m$ if $n$ and $m$ are both odd. That's what "unique" means. The number of factors "$2$" can't be both $1$ and $2$ for the same integer.
A: If we assume that $a,b$ are assumed to be nonnegative integers, then the claim that "both sides of the equation are [positive] integers" will follow.  So let's begin by assuming that.
In this context unique factorization means that (positive) integers can be expressed as a product of (positive) prime integers, with the convention that $1$ is the empty product (one is not considered a prime number in algebra) and that this product is unique except to rearranging the order of the prime numbers (of course $2*3 =3*2$ and similarly).
As a consequence once the prime numbers are collected into like terms, we have unique (positive) exponents on the distinct primes that appear in the factorization.
This allows us to obtain equations for $a,b$ by counting the number of times $7$ and $5$ respectively appear on both sides of the equation.

Even if we were to open the possible solutions $a,b$ to all integers, not necessarily nonnegative ones, it would not change the final conclusion that $a=1,b=2$.  This can be seen by "clearing fractions" (multiplying both sides of the equation by a positive factor that removes denominators).  Then we would be back to both sides being positive integers and applying the previous reasoning.
Indeed the equation shown could be simplified to:
$$ 5^2 * 7^a = 7 * 5^b $$
by cancelling the factors of $2$ and $3$.
Regardless of whether we assume $a,b$ are positive integers, we can invoke unique factorization of integers to show $a=1,b=2$.
